Floor functions stuff

Number Theory Level 3

Find sum of all natural numbers $n$ such that $\left\lfloor \frac { { n }^{ 2 }+n-5 }{ 2 } \right\rfloor$ is a prime.

1 solution

Kushal Bose
Jun 12, 2017

$\dfrac{n^2+n-5}{2}=\dfrac{n^2+n}{2}-2-\dfrac{1}{2}$

As $n^2+n=n(n+1)$ is always even then $\dfrac{n^2+n}{2}$ is always an integer

So, $\lfloor \dfrac{n^2+n-5}{2} \rfloor=\dfrac{n^2+n}{2}-2-1=\dfrac{n^2+n-6}{2}=\dfrac{(n+3)(n-2)}{2}$

The values of $n=3,4$